# What is the equation of the line that is normal to f(x)=sinx-e^xcotx  at  x=pi/3?

Dec 21, 2017

Equation of normal at $x = \frac{\pi}{3}$ is $\left(y - \frac{\sqrt{3}}{2} + {e}^{\frac{\pi}{3}} / \sqrt{3}\right) \left(\frac{1}{2} + \frac{4 - \sqrt{3}}{3} {e}^{\frac{\pi}{3}}\right) = \frac{\pi}{3} - x$

#### Explanation:

Normal is perpendicular to tangent at the pint on the curve.

The slope of tangent to curve $y = f \left(x\right)$ at $x = x +$ is given by $f ' \left({x}_{0}\right)$ and hence slope of normal is $- \frac{1}{f ' \left({x}_{0}\right)}$. Further itpasses through point $\left({x}_{0} , f \left({x}_{0}\right)\right)$ on the curve. This gives us point slope form of normal to the curve as $y - f \left({x}_{0}\right) = - \frac{1}{f ' \left({x}_{0}\right)} \left(x - {x}_{0}\right)$.

Here $y = f \left(x\right) = \sin x - {e}^{x} \cot x$ and ${x}_{0} = \frac{\pi}{3}$

Hence $f \left(\frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right) - {e}^{\frac{\pi}{3}} \cot \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - {e}^{\frac{\pi}{3}} / \sqrt{3}$

Further $f ' \left(x\right) = \cos x + {e}^{x} {\csc}^{2} x - {e}^{x} \cot x$

and $f ' \left(\frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) + {e}^{\frac{\pi}{3}} {\csc}^{2} \left(\frac{\pi}{3}\right) - {e}^{\frac{\pi}{3}} \cot \left(\frac{\pi}{3}\right)$

= $\frac{1}{2} + {e}^{\frac{\pi}{3}} \left[\frac{4}{3} - \frac{1}{\sqrt{3}}\right] = \frac{1}{2} + \frac{4 - \sqrt{3}}{3} {e}^{\frac{\pi}{3}}$

and hence equation of normal is

$y - \frac{\sqrt{3}}{2} + {e}^{\frac{\pi}{3}} / \sqrt{3} = - \frac{1}{\frac{1}{2} + \frac{4 - \sqrt{3}}{3} {e}^{\frac{\pi}{3}}} \left(x - \frac{\pi}{3}\right)$

or $\left(y - \frac{\sqrt{3}}{2} + {e}^{\frac{\pi}{3}} / \sqrt{3}\right) \left(\frac{1}{2} + \frac{4 - \sqrt{3}}{3} {e}^{\frac{\pi}{3}}\right) = \frac{\pi}{3} - x$